0such that kxk ~~for s>,wehave 1 c < Z 0 es h k sds C Z 1 es h k ds D C ye.Cy/ . C y/ ; where y WD h k .Nowwehave .c / c > c y ye.Cy/; since c>. The condition h >c>here plays its role because otherwise C y is not necessarily positive. Notice that the function deﬁned by f.y/ D ye.Cy/ achieves the absolute maximum of e.C1/at y D 1 ,wehave .c / c > c y e.C1/: Correspondingly, we have h < h c C c e.C1/ i k 1 ek C : The righthand side of above inequality is continuous in . Take limit for ! 0,we ﬁnd Œc C c e.C1/ k 1ek C ~~~~for s>,wehave 1 c > Z 0 es h k sds C Z 1 es h k ds D ye.y/ . y/ ; where y WD h k .If y 0,then 86 D. Xie et al. . h/1 ek k ; which is equivalent to h 1 k 1 ek : If y<0then we have 1 c > ye.y/ . y/ D e.y/ 1 y C e.y/ > .y /; which gives h 1 k 1 ek k 1 ek c: Because we are looking for a lower bound, so we take the minimum of the two cases (for ﬁxed >0), and conclude that h 1 k 1 ek k c 1 1 ek : Lemma 6.3.6 h.t/ is continuous for t 2 Œ0; 1/, in particular, limt!0CD c. Proof. Because of Lemma 6.3.1, the only thing left to be justiﬁed is limt!0CD c. For t small, eks D 1 ks,wehave lim t!0C V.h.t/;t/ D lim t!0C e h.t/ k Z t 0 esC h.t/ k eks ds D lim t!0C e h.t/ k Z t 0 esC h.t/ k .1ks/ Because of the continuity and boundedness of h.t/, we can take limit of limt!0C h.t/ inside of the integral and arrive at lim t!0C V.h.t/;t/ D 1 limt!0C h.t/ Compare this with the boundary value of 1 c Œ1ect, we have that limt!0C h.t/ D c. 6 An Asymptotic Method to a Financial Optimization Problem 87 6.4 Numerical Solution of the Free Boundary Since @V @x ¤ 0 we can use Newton method to solve for the free boundary iteratively. Deﬁne Q.h/ D e h k Z t 0 esC h k eks ds 1 c h 1 ect i ; and f.h/D e h k 1 k Z t 0 esC h k eks ds Ce h k Z t 0 esC h k eks 1 k eks ds; our problem is to ﬁnd h such that QŒh.t/ 0; 8t 0: For ﬁxed t DT, discretize Œ0; T uniformly into n subintervals by t0;t1;t2; :::; tn, where t0 D 0; tn DT. Start with h.t0/D c and assume h.t1/; h.t2/; :::; h.tn1/ are known, to compute h.tn/ with Newton’s algorithm, we ﬁrst assign a reasonable initial guess for h.tn/ as h0.tn/D h.tn1/; n D 1I h0.tn/D 2h.tn1/ h.tn2/; n > 1: For a given error tolerance level, say Tole D 107, we have the following Newton’s iteration scheme h.tn/new D h.tn/old Q.h.tn/old / f.h.tn/old/: After each step of iteration, a current error is recorded as error.k/ D h.tn/new h.tn/old : The iteration is kept running until an integer k is reached such that error.k/ < Tole. To increase the accuracy of the numerical solution, one can increase N, the number of grids for partitioning the time interval Œ0; T . For typical parameters with T 25, our numerical simulations show that ND 4096 is large enough for achieving a solution with relative error less than 107, where relative error is deﬁned as the difference of numerical values of h.T /’s achieved with different N’s. Figure 6.1 is a numerical plot of the free boundaries as we ﬁx one set of parameters at a time 88 D. Xie et al. 0 50 100 150 200 0.025 0.03 0.035 0.04 0.045 0.05 t h(t) A Family of the Free Boundary 0 50 100 150 200 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 t h(t) Fig. 6.1 c = 0.05, k D 0:06; 0:07; :::; 0:12 (top to bottom) D 0:06 (right), 0:07 (left). The units for t and h.t/ are years and year1, respectively 6.5 Asymptotic Analysis of the Free Boundary In this section, we derive asymptotic expansions of h.t/ for both small t and large t. Theorem 6.2. As t ! 0, h.t/ c C ˛t,where˛ D.c/k 3 : Proof. We postulate that as t ! 0, h.t/ c C ˛t, plug this into the contract value on h.t/,wehavethat,fort small, V.h.t/;t/ D Z t 0 es cC˛t k Œ1eksds D Z t 0 e .cC˛t/k 2 s2.cC˛t/sds For a; b > 0, s small, we have the following Taylor expansion eas2bs D 1 C as2 C a2s4 2Š C::: 1 bs C b2s2 2Š ::: D 1 bs C a C b2 2 s2 ab C b3 3Š s3 C o.s3/ Integrating it term by term for small t,wehave Z t 0 eas2bsds D t b 2 t2 C 1 3 a C b2 2 t3 1 4 ab C b3 3Š t4 C o.t4/: In terms of our problem at hand, we have 6 An Asymptotic Method to a Financial Optimization Problem 89 V.h.t/;t/ D t c C ˛t 2 t2 C 1 3 hc C ˛t 2 C.c C ˛t/2 2 i t3 o.t3/: We want to match this, term by term, with the known expression of V.h.t/;t/,which is 1 c Œ1 ect D t c 2t2 C 1 3Šc2t3 o.t3/: In the above two series, the coefﬁcients for t and t2 match each other automatically. To match the t3 terms, we need to have ˛ 2 C 1 3 .c /k C c2 2 D 1 6c2; which gives ˛ D.c /k 3 : Theorem 6.3. There exist constants h D limt!1 h.t/, 1 >0, and 2 >0such that, as t !1, h.t/ h 1et; if c<; h.t/ h C 2ect; if c>; where h is implicitly given by M 1; k C 1; h k D c ,whereM.p;q;z/ is the conﬂuent hypergeometric function of the ﬁrst kind of order p , q, and 1 D kc.h /e h k .h c/ ; 2 D k.h / h c : The existence and boundedness of h have been previously shown. The main idea to ﬁnd the exact value of h is to use repeated integration by parts to express the contract value V at t inﬁnity as a inﬁnite series involving h, which turns out to be a conﬂuent hypergeometric function. As in general, given a; b; c > 0,wehave Z 1 0 eayCbecy dy D1 a Z 1 0 ebecy Œeay 0dy D 1 aeb C.1/bc a Z 1 0 e.aCc/yCbecy dy 90 D. Xie et al. Repeating the integration by parts, using the recursive identity Z 1 0 e.aCnc/yCbecy dy D 1 a C ncebC.1/ bc a C nc Z 1 0 e.aC.nC1/y/Cbecy dy; where the tail deﬁnite integral vanishes as n !1,wehave Z 1 0 eayCbecy dy D 1 a nD1X nD1 .1/n bn .a=c C 1/.a=c C 2/:::.a=c C n/eb D 1 aM.1;a=c C 1; b/: In terms of our problems, this means e h k eks Z 1 0 esC h k eks ds D 1 M 1; =k C 1; h k : At t inﬁnity, we want V.h.1/; 1/D e h k eks Z 1 0 esC h k eks ds D 1 c ; which means M 1; k C 1; h k D c : To fully understand the asymptotic behavior of the free boundary as t !1,we evaluate the limit of h0.t/ as t !1. Start with the equation Z 1 0 es h.t/ k Œ1eksds D 1 c h 1 ect i ; take derivative with respect to t along h.t/, h0.t/ k Z t 0 es h.t/ k Œ1eks h 1 eks i ds D et h.t/ k Œ1ekt C ect; we get h0.t/ k D et h.t/ k Œ1ekt ect 1 c Œ1 ect I; 6 An Asymptotic Method to a Financial Optimization Problem 91 where I WD Z t 0 es h.t/ k Œ1ekseksds; which can be evaluated using integration by parts. Thus we get h0.t/ k D .h.t/ / n et h.t/ k Œ1ekt ect o h.t/ c Œ1 ect C et h.t/ k Œ1ekt 1 : When c>, we can write the above equation into h0.t/ D F.t/et; where F.t/ D e h.t/ k Œ1ekt e.c/t h.t/ c.h.t//Œ1 ect C 1 h.t/ n et h.t/ k Œ1ekt o: Since it has been shown that limt!1 h.t/ D h >, it is straightforward to show that F.t/is uniformly bounded in t and limt!1 F.t/D kc.h /e h k h c : Now we postulate h.t/ h 1et; if c>: Compare the limit of h0.t/,weget 1 D kc.h /e h k .h c/ : On the other hand, if c<, we can write the same equation into we can write the above equation into h0.t/ D G.t/ect; where G.t/ D e.c/t h.t/ k Œ1ekt 1 h.t/ c.h.t//Œ1 ect C 1 h.t/ fet h.t/ k Œ1ekt 1g : 92 D. Xie et al. Since it has been shown that limt!1 h.t/ D h ~~